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Levart [38]
3 years ago
12

All modern fertilizers use PO43- as a source of phosphorus. Use your knowledge of the old methods to determine the moles of phos

phorus found in 10.0 grams of (15-15-15)?
Chemistry
1 answer:
tamaranim1 [39]3 years ago
8 0

Answer:

The answer is 0.023 moles of phosphorus

Explanation:

The 15-15-15 fertilizer is a fertilizer of great versatility, made with nitrogen, phosphorus and potassium, which makes it one of the fertilizers most used for fertilizer in the sowing plant, thus covering the crop requirements from planting. .

This fertilizer consists of 14.25% phosphorus pentoxide (P2O5). Therefore, we have to remove 14.25% at 10 grams of 15-15-15 fertilizer to calculate the moles of phosphorus. As follows:

Grams of P2O5 = 10 g x 0.1425 = 1.425 g

We calculate the molecular weight of phosphorus. We use the periodic table:

Phosphorus molecular weight = 2 x 30.97 = 61.94 g/mol

Now we calculate the moles of phosphorus in the fertilizer:

Phosphorus moles = 1,425 g/61.94 g/mol = 0.023 moles

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It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta
Zigmanuir [339]

Answer:

\lambda=241.9\ nm

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that  

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,  

1 mole = 6.023\times 10^{23}\ electrons

So,  

6.023\times 10^{23} electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of \frac {495.0}{6.023\times 10^{23}}\ kJ of energy.

Energy required = 82.18\times 10^{-23}\ kJ

Also,  

1 kJ = 1000 J

So,  

Energy required = 82.18\times 10^{-20}\ J

Also, E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}

\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}

\lambda=\frac{19.878}{10^6\times \:82.18}

\lambda=2.4188\times 10^{-7}\ m

Also,  

1 m = 10⁻⁹ nm

So,  

\lambda=241.9\ nm

6 0
3 years ago
Determine the molecular formula of a compound with an empirical formula of NH2 and a formula mass of 32.06 amu
Lilit [14]
N2H4

<span>Each nitrogen weighs 14.01 and each H weighs 1.01. !4.01+14.01+1.01+1.01 = 32.06 (roughly) </span>

4 0
3 years ago
Read 2 more answers
calculate the volume of air in liters that you might inhale(and exhale)in 8.0 hours. Assume that each breath has a volume of 0.5
maks197457 [2]
Formula:
0.57 liters = 1 breath(17 breaths = 1minute)(60 minutes = 1 hour)(8 hours) = 4651.2 liters

each breath is 0.57 liters
He or she exhales 17(0.57) = 9.69 liters in a minute
in an hour, he or she exhales 9.69(60) liters in a minute = 581.4 liters
Answer = in 8 hours he or she exhales, 581.4(8) = 4651.2 liters in 8 hours


Hope that answers your question!!!!have a great day!!!



6 0
3 years ago
3. What effect does dark energy have on the universe?
sladkih [1.3K]
Dark energy and dark matter seem over analyzed, although they are no different than gravity and wind. Dark energy can be the pulling caused by black holes, stars, and large planets. Dark matter also is just the effect of cosmic dust and debris pushed from explosions but also pulling caused by distant gravitational bodies.
4 0
3 years ago
How many variables should be tested at one time in a scientific investigation?
shutvik [7]
One should be tested at once. You never want to test more because you will mix up your data
6 0
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